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fa-inbox fa-fw post-meta-icon"></i><a class="post-meta-categories" href="/categories/%E6%97%A5%E5%B8%B8/">日常</a></span></div><div class="meta-secondline"><span class="post-meta-separator">|</span><span class="post-meta-wordcount"><i class="far fa-file-word fa-fw post-meta-icon"></i><span class="post-meta-label">字数总计:</span><span class="word-count">1.7k</span><span class="post-meta-separator">|</span><i class="far fa-clock fa-fw post-meta-icon"></i><span class="post-meta-label">阅读时长:</span><span>5分钟</span></span><span class="post-meta-separator">|</span><span class="post-meta-pv-cv" id="" data-flag-title="蓝桥杯2021省赛游记 | C/C++ | 大学A组"><i class="far fa-eye fa-fw post-meta-icon"></i><span class="post-meta-label">阅读量:</span><span id="busuanzi_value_page_pv"></span></span></div></div></div></header><main class="layout" id="content-inner"><div id="post"><article class="post-content" id="article-container"><p>今天想起来还没有写这次比赛的游记，那就写一下吧。</p>
<p>第一次参加蓝桥杯 <s>，因为 ACM 搞不动</s> ，前几天出成绩了——黑龙江省一，还是稍微有点意外的，主要是因为今年蓝桥杯的省赛题简直一言难尽：说它水吧，感觉不比 NOIP 简单，更是比往年的赛题难不少；说它难吧，其实暴力打得好就会有省一。</p>
<h2 id="赛前">赛前</h2>
<p>其实本来不想参加这次比赛的，毕竟性价比比较低，因为身在哈工大，参加此类比赛，花费 300 大洋除了能在简历上加一行，在其他方面如保研等没有任何好处，不像其他大学参加蓝桥杯有各种加分。</p>
<p>不过后来还是决定参加了，因为大一下要专业分流，几乎没有时间卷 ACM，所以还是考一下蓝桥杯保持一下手感吧。</p>
<h2 id="比赛日">比赛日</h2>
<p>这次比赛我选择哈尔滨学院赛点，考前遇到了形如考试机里没有 DevCpp 只有 CodeBlocks 等鬼畜问题，根据规定是可以向组委会投诉的，好在[数据删除]，最终我还是成功用上了 DevCpp 来比赛。</p>
<h3 id="题面">题面</h3>
<p>直接在<a target="_blank" rel="noopener" href="https://blog.csdn.net/ljw_study_in_CSDN/article/details/115836757">这里</a>看吧，就不搬运了。</p>
<h3 id="A-题">A 题</h3>
<p>第一题还是普及组水平的签到题。</p>
<p>直接开一个 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>0</mn><mo>∼</mo><mn>9</mn></mrow><annotation encoding="application/x-tex">0 \sim 9</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">0</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">∼</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">9</span></span></span></span> 的桶，每个桶存 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>2021</mn></mrow><annotation encoding="application/x-tex">2021</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">2</span><span class="mord">0</span><span class="mord">2</span><span class="mord">1</span></span></span></span> 张卡，从 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn></mrow><annotation encoding="application/x-tex">1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span></span></span></span> 开始遍历自然数集，逐数字数位分离，每分离一个数，对应位置的桶减去 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn></mrow><annotation encoding="application/x-tex">1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span></span></span></span> ，直到出现一个数 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi></mrow><annotation encoding="application/x-tex">n</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathnormal">n</span></span></span></span> 的某一位减不动, <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>−</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">n-1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.66666em;vertical-align:-0.08333em;"></span><span class="mord mathnormal">n</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">−</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span></span></span></span> 就是答案。</p>
<p>好像几分钟就秒了这题。</p>
<h3 id="B-题">B 题</h3>
<p>这题我第一反应是类似 NOIP 2017 D2T3 的仪仗队暴力版，不过其实操作会稍微复杂一点， <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>d</mi><mi>p</mi><mo stretchy="false">[</mo><mi>x</mi><mn>1</mn><mo stretchy="false">]</mo><mo stretchy="false">[</mo><mi>y</mi><mn>1</mn><mo stretchy="false">]</mo><mo stretchy="false">[</mo><mi>x</mi><mn>2</mn><mo stretchy="false">]</mo><mo stretchy="false">[</mo><mi>y</mi><mn>2</mn><mo stretchy="false">]</mo></mrow><annotation encoding="application/x-tex">dp[x1][y1][x2][y2]</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathnormal">d</span><span class="mord mathnormal">p</span><span class="mopen">[</span><span class="mord mathnormal">x</span><span class="mord">1</span><span class="mclose">]</span><span class="mopen">[</span><span class="mord mathnormal" style="margin-right:0.03588em;">y</span><span class="mord">1</span><span class="mclose">]</span><span class="mopen">[</span><span class="mord mathnormal">x</span><span class="mord">2</span><span class="mclose">]</span><span class="mopen">[</span><span class="mord mathnormal" style="margin-right:0.03588em;">y</span><span class="mord">2</span><span class="mclose">]</span></span></span></span> 表示 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo stretchy="false">(</mo><mi>x</mi><mn>1</mn><mo separator="true">,</mo><mi>y</mi><mn>1</mn><mo stretchy="false">)</mo><mo separator="true">,</mo><mtext> </mtext><mo stretchy="false">(</mo><mi>x</mi><mn>2</mn><mo separator="true">,</mo><mi>y</mi><mn>2</mn><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">(x1, y1), \ (x2, y2)</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mopen">(</span><span class="mord mathnormal">x</span><span class="mord">1</span><span class="mpunct">,</span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord mathnormal" style="margin-right:0.03588em;">y</span><span class="mord">1</span><span class="mclose">)</span><span class="mpunct">,</span><span class="mspace"> </span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mopen">(</span><span class="mord mathnormal">x</span><span class="mord">2</span><span class="mpunct">,</span><span class="mspace" style="margin-right:0.16666666666666666em;"></span><span class="mord mathnormal" style="margin-right:0.03588em;">y</span><span class="mord">2</span><span class="mclose">)</span></span></span></span> 已经有线连接，遍历所有没有线连接的点元组，然后把属于该直线的所有点对全部标记即可。</p>
<p>虽然做法很简单，但是因为码力衰弱，模拟的时候还出了几个 bug，这题写完时已经快 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>9</mn><mo>:</mo><mn>30</mn></mrow><annotation encoding="application/x-tex">9:30</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">9</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">:</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">3</span><span class="mord">0</span></span></span></span> 了。</p>
<h3 id="C-题">C 题</h3>
<p>这题是傻逼题，不是指题目傻逼，而是我傻逼。</p>
<p>想法很简单，对该数分解质因数，因为看见这个数据规模高达 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><msup><mn>0</mn><mn>16</mn></msup></mrow><annotation encoding="application/x-tex">10^{16}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8141079999999999em;vertical-align:0em;"></span><span class="mord">1</span><span class="mord"><span class="mord">0</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8141079999999999em;"><span style="top:-3.063em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mtight">1</span><span class="mord mtight">6</span></span></span></span></span></span></span></span></span></span></span></span> ，决定采用线筛（事实证明线筛是没有必要的，最暴力的分解方法就可以，毕竟是离线的），最傻逼的事情出现了，我写个线筛写出了这种狗屎代码：</p>
<figure class="highlight cpp"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">1</span>; i &lt;= n; i++)</span><br><span class="line">&#123;</span><br><span class="line">    <span class="keyword">if</span> (!isprime[i])</span><br><span class="line">    &#123;</span><br><span class="line">        prime[++cnt] = i;</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> j = <span class="number">1</span>; i * prime[j] &lt;= n &amp;&amp; j &lt;= cnt; j++)</span><br><span class="line">        &#123;</span><br><span class="line">            isprime[i * prime[j]] = <span class="literal">true</span>;</span><br><span class="line">            <span class="keyword">if</span> (i % prime[j] == <span class="number">0</span>)</span><br><span class="line">                <span class="keyword">break</span>;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<p>错哪就不必多说了，反正考场上我死活没发现哪里错了，手推线筛也推不出来，耗了 30 多 min 还是决定改回埃筛了，分解质因数结果如下：</p>
<p><span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mn>2021041820210418</mn><mo>=</mo><mn>2</mn><mo>×</mo><mn>3</mn><mo>×</mo><mn>3</mn><mo>×</mo><mn>17</mn><mo>×</mo><mn>131</mn><mo>×</mo><mn>2857</mn><mo>×</mo><mn>5882353</mn></mrow><annotation encoding="application/x-tex">2021041820210418=2 \times 3 \times 3 \times 17 \times  131 \times 2857 \times 5882353
</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">2</span><span class="mord">0</span><span class="mord">2</span><span class="mord">1</span><span class="mord">0</span><span class="mord">4</span><span class="mord">1</span><span class="mord">8</span><span class="mord">2</span><span class="mord">0</span><span class="mord">2</span><span class="mord">1</span><span class="mord">0</span><span class="mord">4</span><span class="mord">1</span><span class="mord">8</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:0.72777em;vertical-align:-0.08333em;"></span><span class="mord">2</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">×</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:0.72777em;vertical-align:-0.08333em;"></span><span class="mord">3</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">×</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:0.72777em;vertical-align:-0.08333em;"></span><span class="mord">3</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">×</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:0.72777em;vertical-align:-0.08333em;"></span><span class="mord">1</span><span class="mord">7</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">×</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:0.72777em;vertical-align:-0.08333em;"></span><span class="mord">1</span><span class="mord">3</span><span class="mord">1</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">×</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:0.72777em;vertical-align:-0.08333em;"></span><span class="mord">2</span><span class="mord">8</span><span class="mord">5</span><span class="mord">7</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">×</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">5</span><span class="mord">8</span><span class="mord">8</span><span class="mord">2</span><span class="mord">3</span><span class="mord">5</span><span class="mord">3</span></span></span></span></span></p>
<p>其实接下来的事情就很容易做了，两层 DFS，然后去重完事，但是因为之前的线筛傻逼问题，整得心情很差，不想处理去重这类问题，所以最后没写。</p>
<p>搞完这题时，时间已经来到了 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>10</mn><mo>:</mo><mn>20</mn></mrow><annotation encoding="application/x-tex">10:20</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span><span class="mord">0</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">:</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">2</span><span class="mord">0</span></span></span></span> ，我终究还是太菜。</p>
<h3 id="D-题">D 题</h3>
<p>建图，然后跑最短路完事，我用的还是 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>D</mi><mi>i</mi><mi>j</mi><mi>k</mi><mi>s</mi><mi>t</mi><mi>r</mi><mi>a</mi></mrow><annotation encoding="application/x-tex">Dijkstra</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8888799999999999em;vertical-align:-0.19444em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">D</span><span class="mord mathnormal">i</span><span class="mord mathnormal" style="margin-right:0.05724em;">j</span><span class="mord mathnormal" style="margin-right:0.03148em;">k</span><span class="mord mathnormal">s</span><span class="mord mathnormal">t</span><span class="mord mathnormal" style="margin-right:0.02778em;">r</span><span class="mord mathnormal">a</span></span></span></span> ，尽管这么小的数据规模用 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>F</mi><mi>l</mi><mi>o</mi><mi>y</mi><mi>d</mi></mrow><annotation encoding="application/x-tex">Floyd</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8888799999999999em;vertical-align:-0.19444em;"></span><span class="mord mathnormal" style="margin-right:0.13889em;">F</span><span class="mord mathnormal" style="margin-right:0.01968em;">l</span><span class="mord mathnormal">o</span><span class="mord mathnormal" style="margin-right:0.03588em;">y</span><span class="mord mathnormal">d</span></span></span></span> 随便跑（我是傻逼）。主要问题在于我不好造数据来测试自己的最短路有没有写挂，要是写挂了，因为这是填空题，那就 gg 了。 <s>这么看来，洛谷的最短路模板题给的样例数据质量还是不错的。</s></p>
<h3 id="E-题">E 题</h3>
<p>哈密顿回路计数不记得怎么写了，就先跳了。</p>
<p>最后半个小时回来写了个暴力，发现 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>=</mo><mn>15</mn></mrow><annotation encoding="application/x-tex">n=15</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathnormal">n</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span><span class="mord">5</span></span></span></span> 时已经要跑 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>11</mn><mi>s</mi></mrow><annotation encoding="application/x-tex">11s</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span><span class="mord">1</span><span class="mord mathnormal">s</span></span></span></span> 才能跑出结果， <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi></mrow><annotation encoding="application/x-tex">n</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathnormal">n</span></span></span></span> 每增加 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn></mrow><annotation encoding="application/x-tex">1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span></span></span></span> ，时间复杂度增加一个数量级这样，题目要求 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>=</mo><mn>21</mn></mrow><annotation encoding="application/x-tex">n=21</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.43056em;vertical-align:0em;"></span><span class="mord mathnormal">n</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">2</span><span class="mord">1</span></span></span></span> 的结果，那还是算辽，跑不出来告辞。</p>
<h3 id="F-题">F 题</h3>
<p>第一道编程题。</p>
<p>乍一看以为是 NOIP 2018 D1T2 的货币系统，可以用类似 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>d</mi><mi>p</mi></mrow><annotation encoding="application/x-tex">dp</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8888799999999999em;vertical-align:-0.19444em;"></span><span class="mord mathnormal">d</span><span class="mord mathnormal">p</span></span></span></span> 的方式标记所有可能的重量，这题多了个减法的要求，所以需要稍微处理一下防止同一个砝码使用多次。</p>
<h3 id="G-题">G 题</h3>
<p>博弈论不会写。主要是因为我不知道什么样的策略算最优，又没有部分分，就不写了。</p>
<h3 id="H-题">H 题</h3>
<p>比较有意思的题，第一次听说所谓“左孩子右兄弟”的构造方法。一开始没看懂题，看了一下样例才懂。赛后我同<a target="_blank" rel="noopener" href="https://laybxc.github.io/">laybxc</a>解释样例的时候使用了如下例子：</p>
<p><img src= "" data-lazy-src="https://z3.ax1x.com/2021/04/30/gEgw6O.md.jpg" alt="gEgw6O.md.jpg"></p>
<p>稍微思考一下就会发现，最优的构造策略是，对于每个节点，把深度贡献最大的子节点放到左子树的末端就可以了，这样每个节点的贡献就是：</p>
<p><span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mtext>子节点数</mtext><mo>+</mo><mtext>贡献最大的子节点的贡献</mtext></mrow><annotation encoding="application/x-tex">子节点数+贡献最大的子节点的贡献
</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.76666em;vertical-align:-0.08333em;"></span><span class="mord cjk_fallback">子</span><span class="mord cjk_fallback">节</span><span class="mord cjk_fallback">点</span><span class="mord cjk_fallback">数</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222222222222222em;"></span></span><span class="base"><span class="strut" style="height:0.68333em;vertical-align:0em;"></span><span class="mord cjk_fallback">贡</span><span class="mord cjk_fallback">献</span><span class="mord cjk_fallback">最</span><span class="mord cjk_fallback">大</span><span class="mord cjk_fallback">的</span><span class="mord cjk_fallback">子</span><span class="mord cjk_fallback">节</span><span class="mord cjk_fallback">点</span><span class="mord cjk_fallback">的</span><span class="mord cjk_fallback">贡</span><span class="mord cjk_fallback">献</span></span></span></span></span></p>
<p>写法有点类似树链剖分，两次 DFS 可以统计出答案。</p>
<h3 id="I-题">I 题</h3>
<p>记得这种题见过不止一次，但是这次确实不会写了。思考了几种算法，大多是 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>d</mi><mi>p</mi></mrow><annotation encoding="application/x-tex">dp</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8888799999999999em;vertical-align:-0.19444em;"></span><span class="mord mathnormal">d</span><span class="mord mathnormal">p</span></span></span></span> ，但是形如 <code>))))(((((((((</code> 这样的样例依然没有想到一个很好的办法处理。</p>
<h3 id="F-题-2">F 题</h3>
<p>没时间了，题目也没看，虽然有部分分可以骗，\。</p>
<h2 id="赛后">赛后</h2>
<p>其实考试的时候我一直看的是机房电脑的时间，但是它比实际时间慢 10min，所以我以为是 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>12</mn><mo>:</mo><mn>50</mn></mrow><annotation encoding="application/x-tex">12:50</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">1</span><span class="mord">2</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span><span class="mrel">:</span><span class="mspace" style="margin-right:0.2777777777777778em;"></span></span><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">5</span><span class="mord">0</span></span></span></span> 的时候就突然收卷了，虽然我已经写完了。</p>
<p>期望得分 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>55</mn></mrow><annotation encoding="application/x-tex">55</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">5</span><span class="mord">5</span></span></span></span> 吧，主要是今年蓝桥杯画风也挺鬼畜的，有 NOIP 2018 的感觉了，填空题还是挺刺激的，如果以后有机会，来打打也不是不行。</p>
<p>回校的路上和<a target="_blank" rel="noopener" href="https://www.goldenpotato.cn/">GoldenPotato</a>对了一下答案，填空题我写了的部分都是一样的，感觉还行，不过赛后期望分数和赛前期望的分数相比差距还是不小的，毕竟低估了比赛难度。</p>
<p>最终得了省一，还是相对满意的，不过似乎被其他学校暴打了，尽管这次比赛哈工大只来了 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>3</mn></mrow><annotation encoding="application/x-tex">3</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">3</span></span></span></span> 人——我，<a target="_blank" rel="noopener" href="https://www.goldenpotato.cn/">GoldenPotato</a>，和另一位未知老哥。好在 <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>300</mn></mrow><annotation encoding="application/x-tex">300</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.64444em;vertical-align:0em;"></span><span class="mord">3</span><span class="mord">0</span><span class="mord">0</span></span></span></span> 大洋没被浪费。</p>
</article><div class="post-copyright"><div class="post-copyright__author"><span class="post-copyright-meta">文章作者: </span><span class="post-copyright-info"><a href="mailto:undefined">Von Brank</a></span></div><div class="post-copyright__type"><span class="post-copyright-meta">文章链接: </span><span class="post-copyright-info"><a href="https://vonbrank.github.io/archives/lanqiao-cup-2021-province-experience/">https://vonbrank.github.io/archives/lanqiao-cup-2021-province-experience/</a></span></div><div class="post-copyright__notice"><span class="post-copyright-meta">版权声明: </span><span class="post-copyright-info">本博客所有文章除特别声明外，均采用 <a href="https://creativecommons.org/licenses/by-nc-sa/4.0/" target="_blank">CC BY-NC-SA 4.0</a> 许可协议。转载请注明来自 <a href="https://vonbrank.github.io" target="_blank">Von Brank</a>！</span></div></div><div class="tag_share"><div class="post-meta__tag-list"><a class="post-meta__tags" href="/tags/%E6%97%A5%E5%B8%B8/">日常</a></div><div class="post_share"><div 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